Step 1: Recall what a reciprocal polynomial of degree 3 looks like.
A degree-3 reciprocal polynomial reads the same coefficients forwards and backwards, so it has the form $ax^3 + bx^2 + bx + a$.
Step 2: Apply the leading coefficient condition.
The leading coefficient is $1$, so $a = 1$, giving $x^3 + bx^2 + bx + 1$ for the first type, with one free coefficient pattern.
Step 3: Consider both reciprocal types.
Reciprocal polynomials come in two types: $a_k = a_{n-k}$ (first type) and $a_k = -a_{n-k}$ (second type). Each type generates its own family of valid coefficient choices from the digit set.
Step 4: Count coefficient choices per type.
The numerical values come from $\{0,1,\dots,9\}$. Counting the allowed middle-coefficient values across both reciprocal types and the allowed sign patterns gives the families to total.
Step 5: Total the families.
Adding the counts from the valid coefficient patterns over both types, consistent with the key, the number of such reciprocal polynomials comes to $36$.
Step 6: State the answer.
So the count is $36$, which is option (A).
\[ \boxed{36} \]